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Genetic 342 A&B

Kenneth G Wilson

Professory of Botany

Trihybrid cross Assuming Independent Assortment

There are two possible solutions

Classical Solution

In Drosophila the following wild type alleles are used for analysis v (vermilion eyes), cv (cross-veinless) and ct (cut or snipped wing edges). Flies doubly recessive and homozygous (v+v+,cvcv,ctct) are crossed with flies singly recessive and homozygous (vv, cv+cv+, ct+ct+). Female progeny that are vv+, cvcv+, ctct+ are then test crossed with the tripley recessive males (vv, cvcv, ctct)

P1=v+v+,cvcv,ctct x vv, cv+cv+, ct+ct+

F1 =Wild Type - V+V, cv+cv, ct+ct x vv, cvcv, ctct (gametes v, cv, ct)

F1 gamete types Obs. Description

v

cv+

ct+

580

Parental

v+

cv

ct

592

Parental

v

cv

ct+

45

Crossover Region I

v+

cv+

ct

40

Crossover Region I

v

cv

ct

89

Crossover Region II

v+

cv+

ct+

94

Crossover Region II

v

cv+

ct

3

Double Crossover

v+

cv

ct+

5

Double Crossover

     

1448

 

Finding the center gene by comparing parental to double crossover.

580+59 2

Parental

45+40

Crossover Region I

89+94

Crossover Region II

3+5

Double Crossover

1448

 

 

C/O

Chromosome region

% Recombination

45+40

Crossover Region I

[(45+40)+(3+5)]x100/1448=6.4%

89+94

Crossover Region II

[(89+94)+(3+5)]x100/1448=13.2%

3+5

Double Crossover

 
     

 

Map data
v<---13.2--->ct<--6.4-->cv

 

Can we calculate the Interference?
   The double cross overs are always the smallest in number. In this problem that is 5+3.   The expected frequency of double crossovers can be calculated by multiplying the   frequency of a crossover between v and ct by the frequency of a crossover between ct  and cv. 0.132 x 0.064 = 0.0084

The expected number of double crossovers is equal to the expected frequence of  double crossovers times the total number of individuals . 0.0084 x 1448 = 12

   Interference (I) = 1 -([# of dble c/o]/[exp # of dble c/o]) = 1- (8/12) = 0.33 }

oliver

This is our friend Oliver the parrotlet,

Oliver is a true parrot,

the smallest in the world.